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Math Help - finding length

  1. #1
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    Question finding length

    A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?
    Last edited by zak_killcity14@hotmail.co; December 1st 2006 at 06:50 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by zak_killcity14@hotmail.co View Post
    A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?
    the figure is like this:
    make a right angled triangle, right angled at D
    label one vertex as A, other as B
    On DB, choose a point C
    Now, angleABD(or ABC)=45degrees
    and angleACD=67degrees
    BC=32 m
    you have to find AD
    from triangle ACD
    \frac{AD}{CD}=tan67
    or CD=\frac{AD}{tan67}......(1)
    from triangle ABD
    \frac{AD}{BD}=tan45
    or BD=\frac{AD}{tan45}......(2)
    Subtracting (1) from (2)
    BD-CD=\frac{AD}{tan45} - \frac{AD}{tan67}
    but BD-CD=BC=32
    hence
    32=\frac{AD}{tan45} - \frac{AD}{tan67}
    AD= \frac{32}{\frac{1}{tan45} - \frac{1}{tan67}}

    Keep Smiling
    Malay
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  3. #3
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    Hello, Zak!

    I assume you have a sketch . . .


    A steel chimney is held in position by several guy wires.
    Two such wires AB and AC where B,\,C and the base of the chimney, D,
    are on the same level in a straight line with B and C on the left of the chimney.
    If \text{angle }ABC = 45^o,\text{ angle }ACD = 67^o,\;BC = 32\text{ m,}
    find the height of the chimney, AD.
    Code:
                                  * A
                               * *|
                            *   * |
                         *     *  |
                      *       *   | y
                   *         *    |
                *           *     |
             * 45         * 67  |
          *---------------*-------*
          B      32       C   x   D

    Let x = CD,\:y = AD.

    In right triangle ADB:\;\tan45^o = \frac{y}{x+32}\quad\Rightarrow\quad1 = \frac{y}{x+32}\quad\Rightarrow\quad x \:= \:y - 32 [1]

    In right triangle ADC:\;\tan67^o = \frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan67^o} [2]


    Equate [1] and [2]: . y - 32 \:=\:\frac{y}{\tan67^o}


    Solve for y:\;\;y\tan67^o - 32\tan67^o \:=\:y\quad\Rightarrow\quad y\tan67^o  - 32\tan67^o \:=\:y

    . . y\tan67^o - y \:=\:32\tan67^o\quad \Rightarrow\quad y(\tan67^o - 1) \:=\:32\tan67^o


    Therefore: . y \;=\;\frac{32\tan67^o}{\tan67^o - 1} \;=\;55.60138965 \;\approx\;\boxed{55.6\text{ m}}

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